A wire of length L₀ is supplied heat to raise its temperature by T. If γ is the coefficient of volume expansion of the wire and Y is the Young’s modulus of the wire then the energy density stored in the wire is

The correct option is $\frac{{\gamma }^2{T}^{2}Y}{18}$.

To find the energy density stored in the wire due to thermal expansion, we can break the problem down into the following steps:

Step 1: Relate the coefficient of volume expansion to the coefficient of linear expansion
The coefficient of volume expansion ($\gamma$) of a material is related to its coefficient of linear expansion ($\alpha$) by the relationship:
$\gamma =3\alpha$
Rearranging this equation gives the coefficient of linear expansion:
$\alpha =\frac{\gamma }{3}$

Step 2: Determine the thermal strain
When the temperature of the wire is raised by $T$, it attempts to expand. The thermal strain ($\epsilon$) produced in the wire is given by:
$\epsilon =\alpha T$
Substituting the value of $\alpha$ from Step 1:
$\epsilon =\frac{\gamma }{3}T$

Step 3: Calculate the energy density
The elastic potential energy stored per unit volume (energy density, $u$) is given by the formula:
$u=\frac{1}{2}\times \text{Stress}\times \text{Strain}$
According to Hooke's Law, $\text{Stress}=Y\times \text{Strain}$, where $Y$ is the Young's modulus of the wire. Therefore, we can write:
$u=\frac{1}{2}Y{\epsilon }^2$

Now, substituting the thermal strain expression ($\epsilon =\frac{\gamma T}{3}$) into the equation:
$u=\frac{1}{2}Y{\left(\frac{\gamma T}{3}\right)}^2$
$u=\frac{1}{2}Y\left(\frac{{\gamma }^2{T}^2}{9}\right)$
$u=\frac{{\gamma }^2{T}^{2}Y}{18}$